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Why is it true that $intint f(x-t)g(t)dtdx = int f(x)dx int g(x)dx$?
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This appears in a proof on page 122 of Measure and Integral by Wheeden and Zygmund. The theorem is that
If $f,g$ are Lebesgue integrable functions from $mathbb R^n$ to $mathbb R$, then $(f*g)(x)$ exists for almost every $mathbf xin mathbb R^n$ and is measurable. Morevover, $f*g$ is Lebesgue integrable and$$int_mathbb R^n |f*g|dmathbf x le left(int_mathbb R^n |f|dmathbf xright)left(int_mathbb R^n |g|dmathbf xright),$$ $$int_mathbb R^n (f*g)dmathbf x le left(int_mathbb R^n fdmathbf xright)left(int_mathbb R^n gdmathbf xright).$$
On the last few line of the proof, they say by Fubini's theorem, $int f(mathbf x -mathbf t)g(mathbf t) dmathbf t$ exists for a.e. $mathbf x$ and is measurable and integrable; also, $$intint f(x-t)g(t)dtdx = int left[int f(x-t)g(t)dt right] dx = int f(x)dx int g(x)dx.$$
Why does the last equality hold? I guessed they use change of variable? How do they change the variable?
real-analysis proof-explanation
asked Jul 25 at 21:39
user398843
394215
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up vote
2
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This appears in a proof on page 122 of Measure and Integral by Wheeden and Zygmund. The theorem is that
If $f,g$ are Lebesgue integrable functions from $mathbb R^n$ to $mathbb R$, then $(f*g)(x)$ exists for almost every $mathbf xin mathbb R^n$ and is measurable. Morevover, $f*g$ is Lebesgue integrable and$$int_mathbb R^n |f*g|dmathbf x le left(int_mathbb R^n |f|dmathbf xright)left(int_mathbb R^n |g|dmathbf xright),$$ $$int_mathbb R^n (f*g)dmathbf x le left(int_mathbb R^n fdmathbf xright)left(int_mathbb R^n gdmathbf xright).$$
On the last few line of the proof, they say by Fubini's theorem, $int f(mathbf x -mathbf t)g(mathbf t) dmathbf t$ exists for a.e. $mathbf x$ and is measurable and integrable; also, $$intint f(x-t)g(t)dtdx = int left[int f(x-t)g(t)dt right] dx = int f(x)dx int g(x)dx.$$
Why does the last equality hold? I guessed they use change of variable? How do they change the variable?
real-analysis proof-explanation
asked Jul 25 at 21:39
user398843
394215
2
Change the order of integration. $int f(x-t),dx = int f(x),dx$ by the translation-invariance of the Lebesgue measure.
– Daniel Fischer♦
Jul 25 at 21:43
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
This appears in a proof on page 122 of Measure and Integral by Wheeden and Zygmund. The theorem is that
If $f,g$ are Lebesgue integrable functions from $mathbb R^n$ to $mathbb R$, then $(f*g)(x)$ exists for almost every $mathbf xin mathbb R^n$ and is measurable. Morevover, $f*g$ is Lebesgue integrable and$$int_mathbb R^n |f*g|dmathbf x le left(int_mathbb R^n |f|dmathbf xright)left(int_mathbb R^n |g|dmathbf xright),$$ $$int_mathbb R^n (f*g)dmathbf x le left(int_mathbb R^n fdmathbf xright)left(int_mathbb R^n gdmathbf xright).$$
On the last few line of the proof, they say by Fubini's theorem, $int f(mathbf x -mathbf t)g(mathbf t) dmathbf t$ exists for a.e. $mathbf x$ and is measurable and integrable; also, $$intint f(x-t)g(t)dtdx = int left[int f(x-t)g(t)dt right] dx = int f(x)dx int g(x)dx.$$
Why does the last equality hold? I guessed they use change of variable? How do they change the variable?
real-analysis proof-explanation
asked Jul 25 at 21:39
user398843
394215
This appears in a proof on page 122 of Measure and Integral by Wheeden and Zygmund. The theorem is that
If $f,g$ are Lebesgue integrable functions from $mathbb R^n$ to $mathbb R$, then $(f*g)(x)$ exists for almost every $mathbf xin mathbb R^n$ and is measurable. Morevover, $f*g$ is Lebesgue integrable and$$int_mathbb R^n |f*g|dmathbf x le left(int_mathbb R^n |f|dmathbf xright)left(int_mathbb R^n |g|dmathbf xright),$$ $$int_mathbb R^n (f*g)dmathbf x le left(int_mathbb R^n fdmathbf xright)left(int_mathbb R^n gdmathbf xright).$$
On the last few line of the proof, they say by Fubini's theorem, $int f(mathbf x -mathbf t)g(mathbf t) dmathbf t$ exists for a.e. $mathbf x$ and is measurable and integrable; also, $$intint f(x-t)g(t)dtdx = int left[int f(x-t)g(t)dt right] dx = int f(x)dx int g(x)dx.$$
Why does the last equality hold? I guessed they use change of variable? How do they change the variable?
real-analysis proof-explanation
asked Jul 25 at 21:39
user398843
394215
edited Jul 25 at 21:55
asked Jul 25 at 21:39
user398843
394215
asked Jul 25 at 21:39
user398843
394215
asked Jul 25 at 21:39
user398843
394215
394215
2
Change the order of integration. $int f(x-t),dx = int f(x),dx$ by the translation-invariance of the Lebesgue measure.
– Daniel Fischer♦
Jul 25 at 21:43
add a comment |Â
2
Change the order of integration. $int f(x-t),dx = int f(x),dx$ by the translation-invariance of the Lebesgue measure.
– Daniel Fischer♦
Jul 25 at 21:43
2
2
Change the order of integration. $int f(x-t),dx = int f(x),dx$ by the translation-invariance of the Lebesgue measure.
– Daniel Fischer♦
Jul 25 at 21:43
Change the order of integration. $int f(x-t),dx = int f(x),dx$ by the translation-invariance of the Lebesgue measure.
– Daniel Fischer♦
Jul 25 at 21:43
add a comment |Â
2 Answers
2
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up vote
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accepted
It should be
$$int int f(x-t) g(t) , dt , dx = int left( int f(x-t)g(t) , dx right) dt = int left( int f(x-t) , dx right) g(t) , dt, $$
because $g(t)$ is independent of $x$. Lebesgue measure is translation-invariant, so $int f(x-t) , dx = int f(x) , dx $, and then this independent of $t$, so
$$ int left( int f(x) , dx right) g(t) , dt = left( int f(x) , dx right) int g(t) , dt $$
as desired.
answered Jul 25 at 21:45
Chappers
55k74190
add a comment |Â
up vote
1
down vote
Use Fubini to express it as
$$
int g(t)left( int f(x-t) dx right) dt.
$$
Then because Lebesgue measure is translation invariant, the bracket is independent of $t$ and coincides with $int f(x) dx$. Pull it out and you arrive at the desired expression.
answered Jul 25 at 21:45
Jan Bohr
3,0991419
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
It should be
$$int int f(x-t) g(t) , dt , dx = int left( int f(x-t)g(t) , dx right) dt = int left( int f(x-t) , dx right) g(t) , dt, $$
because $g(t)$ is independent of $x$. Lebesgue measure is translation-invariant, so $int f(x-t) , dx = int f(x) , dx $, and then this independent of $t$, so
$$ int left( int f(x) , dx right) g(t) , dt = left( int f(x) , dx right) int g(t) , dt $$
as desired.
answered Jul 25 at 21:45
Chappers
55k74190
add a comment |Â
up vote
4
down vote
accepted
It should be
$$int int f(x-t) g(t) , dt , dx = int left( int f(x-t)g(t) , dx right) dt = int left( int f(x-t) , dx right) g(t) , dt, $$
because $g(t)$ is independent of $x$. Lebesgue measure is translation-invariant, so $int f(x-t) , dx = int f(x) , dx $, and then this independent of $t$, so
$$ int left( int f(x) , dx right) g(t) , dt = left( int f(x) , dx right) int g(t) , dt $$
as desired.
answered Jul 25 at 21:45
Chappers
55k74190
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
It should be
$$int int f(x-t) g(t) , dt , dx = int left( int f(x-t)g(t) , dx right) dt = int left( int f(x-t) , dx right) g(t) , dt, $$
because $g(t)$ is independent of $x$. Lebesgue measure is translation-invariant, so $int f(x-t) , dx = int f(x) , dx $, and then this independent of $t$, so
$$ int left( int f(x) , dx right) g(t) , dt = left( int f(x) , dx right) int g(t) , dt $$
as desired.
answered Jul 25 at 21:45
Chappers
55k74190
It should be
$$int int f(x-t) g(t) , dt , dx = int left( int f(x-t)g(t) , dx right) dt = int left( int f(x-t) , dx right) g(t) , dt, $$
because $g(t)$ is independent of $x$. Lebesgue measure is translation-invariant, so $int f(x-t) , dx = int f(x) , dx $, and then this independent of $t$, so
$$ int left( int f(x) , dx right) g(t) , dt = left( int f(x) , dx right) int g(t) , dt $$
as desired.
answered Jul 25 at 21:45
Chappers
55k74190
answered Jul 25 at 21:45
Chappers
55k74190
answered Jul 25 at 21:45
Chappers
55k74190
answered Jul 25 at 21:45
Chappers
55k74190
55k74190
add a comment |Â
add a comment |Â
up vote
1
down vote
Use Fubini to express it as
$$
int g(t)left( int f(x-t) dx right) dt.
$$
Then because Lebesgue measure is translation invariant, the bracket is independent of $t$ and coincides with $int f(x) dx$. Pull it out and you arrive at the desired expression.
answered Jul 25 at 21:45
Jan Bohr
3,0991419
add a comment |Â
up vote
1
down vote
Use Fubini to express it as
$$
int g(t)left( int f(x-t) dx right) dt.
$$
Then because Lebesgue measure is translation invariant, the bracket is independent of $t$ and coincides with $int f(x) dx$. Pull it out and you arrive at the desired expression.
answered Jul 25 at 21:45
Jan Bohr
3,0991419
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Use Fubini to express it as
$$
int g(t)left( int f(x-t) dx right) dt.
$$
Then because Lebesgue measure is translation invariant, the bracket is independent of $t$ and coincides with $int f(x) dx$. Pull it out and you arrive at the desired expression.
answered Jul 25 at 21:45
Jan Bohr
3,0991419
Use Fubini to express it as
$$
int g(t)left( int f(x-t) dx right) dt.
$$
Then because Lebesgue measure is translation invariant, the bracket is independent of $t$ and coincides with $int f(x) dx$. Pull it out and you arrive at the desired expression.
answered Jul 25 at 21:45
Jan Bohr
3,0991419
answered Jul 25 at 21:45
Jan Bohr
3,0991419
answered Jul 25 at 21:45
Jan Bohr
3,0991419
answered Jul 25 at 21:45
Jan Bohr
3,0991419
3,0991419
add a comment |Â
add a comment |Â
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2
Change the order of integration. $int f(x-t),dx = int f(x),dx$ by the translation-invariance of the Lebesgue measure.
– Daniel Fischer♦
Jul 25 at 21:43